You are sitting in your car at a traffic light at the foot of a 200 ft long ramp that will take you onto the interstate. If your cat can at best accelerate at 5 m/s^2 and the traffic on the interstate is moving at 65mph. will the ramp be long enough for you to reach that speed and thus safely merge with the traffic?

200ft=60.96m
a=5m/s^2
v=65mph=20.95m/s
v_0=0m/s

would I use delta d=(v^2 - v_0^2)/(2a)
60.96=29.05^2-0/(2*5)
60.96=84.39
so thus the answer would be no the car would not safely merge with the traffic. Is this correct? Thanks in advance for your help.

correct.

that is one fast-moving cat.

To determine if the ramp is long enough for you to reach the required speed and safely merge with the traffic, you can use the equation for displacement:

Δd = (v^2 - v0^2) / (2a)

where Δd is the displacement, v is the final velocity, v0 is the initial velocity, and a is the acceleration.

In this case, the initial velocity (v0) is 0 m/s, the final velocity (v) is 20.95 m/s, and the acceleration (a) is 5 m/s^2. The displacement (Δd) represents the length of the ramp, which is 60.96 m.

Plugging in the values into the equation, we have:

Δd = (20.95^2 - 0) / (2 * 5)
60.96 = 436.82 / 10
60.96 = 43.682

Based on the calculation, it seems there was an error in your calculations. The correct value for Δd is 43.682 meters, not 60.96 meters.

Therefore, the ramp would indeed be long enough for you to reach the required speed and safely merge with the traffic.